AM - What's really happening?

[k4kyv]

Side bands exist on paper, and they exist in reality, as proof I offer that I have tuned through them with a narrow filter on the receiver.

But try doing that with a crystal set.  As far as it, or an oscilloscope with the deflection plates tied directly to the pickup antenna is concerned, it's a single carrier of varying amplitude.

Is it wrong to simplify this and say that the upper sideband might be produced by the positive swing of the audio cycle and the lower sideband might be produced by the negative swing, except that the RF frequency is so high that this relationship is reversed every half cycle of the RF wave?

That is absolutely wrong.  The upper and lower sidebands are produced through the sum and difference heterodyning process.  They have nothing to do with positive and negative modulation peaks.

I recall an early issue of QST when slopbucket was first being presented to the mainstream amateur community, in which there was a picture on the front cover that conveyed that same false notion, and as a result I have had people argue with me till they were blue in the face that the diode detector in a receiver converts an AM signal to SSB, because it rectifies half of the envelope of the modulated carrier as it "swings" through the modulation cycle.

If that proposition were true, we wouldn't need those expensive crystal or mechanical selectivity filters to generate SSB.  A simple diode rectifier would do the trick.

[Gito]

Hi


How can the amplifier conti...to deliver the always present carrier....

Maybe because when you modulate the transmitter it's a combination of carrier +modulation with  positive and negative signal .The scope can only "see" the combination of it

When you have a 100 volt carrier modulated it with a sine wave say 50 v Rms,you don't only modulate 50 v negative but also 50 v positive

so  100 v (carrier) -  50 v(negative cycle) + 50 v(positive cycle) = 100 v.

So the carrier amplitude level stay the same seen by the scope.

don't laught at me if I am wrong .Just an idea

Regards


Gito

[Opcom]

Side bands exist on paper, and they exist in reality, as proof I offer that I have tuned through them with a narrow filter on the receiver.

But try doing that with a crystal set.  As far as it, or an oscilloscope with the deflection plates tied directly to the pickup antenna is concerned, it's a single carrier of varying amplitude.

Is it wrong to simplify this and say that the upper sideband might be produced by the positive swing of the audio cycle and the lower sideband might be produced by the negative swing, except that the RF frequency is so high that this relationship is reversed every half cycle of the RF wave?

That is absolutely wrong.  The upper and lower sidebands are produced through the sum and difference heterodyning process.  They have nothing to do with positive and negative modulation peaks.

I recall an early issue of QST when slopbucket was first being presented to the mainstream amateur community, in which there was a picture on the front cover that conveyed that same false notion, and as a result I have had people argue with me till they were blue in the face that the diode detector in a receiver converts an AM signal to SSB, because it rectifies half of the envelope of the modulated carrier as it "swings" through the modulation cycle.

If that proposition were true, we wouldn't need those expensive crystal or mechanical selectivity filters to generate SSB.  A simple diode rectifier would do the trick.

not the modulation peaks. The modulation, as in the direction and speed a modulating voltage is changing, added to or multiplied with, the direction and speed of change of another voltage the carrier. The modulation voltage is headed + or - at a certain rate, just like the carrier is. The product of the rates of change determine how far out the sideband is from the carrier. It's what I was visualizing with the vectors.

[k4kyv]

That sounds more like the way I visualise phase modulation that occurs with frequency modulation and vice versa, but I can't see any relationship to LSB and USB with AM.  I would say that both the positive and negative swings of the audio cycle each produce both USB and LSB components, since each of the two sidebands of an AM signal is an exact mirror image of the other.  This is not the case with the sideband components of an FM signal.

[Gito]

HI

Chas stated--  How can------the always present carrier power  when its plate is Zero?

Is it true what Chas wrote?
in the Radio Communication hand Handbook (RSGB )
I have read in Modulation Systems *chapter 9)

.......When the negative peak of the modulation signal has reduced the amplitude of the CARRIER TO ZERO........so no carrier no output I think,and no carrier power output.

That means when we modulate t0 100% ,at peak of negative modulation
the Carrier output is Zero.
but the Modulation proses does not stop there,so we always get a carrier  because the positive and negative modulation,but  in the proses there are times when the Carrier output is Zero ,depending of the amplitude of the negative signal .

Regards

Gito

[kc6mcw]

Its because of the flywheel effect...right? It replaces the carrier back into the voids.

[Gito]

I don't know if it's a flywheel effect but what I know after the negative cycle modulation it's followed  by a positive modulation.
(assuming it"s a sine wave modulation)
The point is if the Plate voltage is zero ,because the negative modulation ,at that time there's no carrier output.

Regards

Gito.

[W3RSW]

Yes, AM. It's called amplitude modulation of the carrier, from four or greater (depending on voice characteristics) times the peak envelope power of the unmodulated carrier to zero. 

Absolutely, there are times (instantaneously small in a 100%, correctly modulated signal) when the carrier, for that matter the whole signal is zero.

As you exceed, just ever so slightly, 100% modulation of a distortion free sine wave, you'll see zero signal, power, everything.

 The instantaneous crossing into the region of zero power, of course, generates distortion, just like any square wave or very abrubt transistion of current.  Luckily it's at a very low power level. It is an indicator, usually, of much greater distortion in the rest of the signal, given the way such signals are generated by the general population of radio amateurs. 

[kc6mcw]

Is it safe to assume that the actual carrier NEVER goes to zero by offering the example of a SSB station talking to another SSB station on a particular frequency. While listening to them on AM, they would sound scrambled of course. But then when someone else on frequency transmits an AM carrier, you would then hear the SSB stations clearly due to the AM carrier being re-inserted. Would that be proof that the carrier is solid all the time and never goes to zero? It just shows us that on the scope, but the scope shows us an instantaneous display of a frequency even if that frequency is BEATING with another frequency therefore showing us a new strange looking frequency... So when it appears that the carrier is going to zero, it may not be the case in the real world.

[WB6VHE]

Yes, Rick is correct.  The signal goes to zero, carrier included!  Just look at your scope with
an envelope pattern displayed: sig is zero on neg peaks at 100%.  Period.  I think people need to re-read
post #45 from 2005.  I've seen this confusion in courses I have taught, and it seems to arise
from misunderstanding the frequency domain versus time domain.  I tried to explain this in
my previous post, but maybe I didn't do a good enough job!  Try this:  consider a 1 kHz sinusoidal
audio tone.  The amplitude clearly goes through zero every 1/2 cycle, but you don't hear it doing so:
the tone is continuous.  This is the difference between the time domain (sig viewed on o'scope)
and the frequency domain (your ear operates as a detector in the frequency domain, just like
your receiver or spectrum analyzer).  The ear integrates the signal over time, just like the
spectrum analyzer.  After integration, time is no longer a parameter: it has been integrated out
(Fourier transformed out...when you go from time domain to frequency domain, it's a Fourier
transform and the signal has been integrated over time, so time is out of the picture, so to speak,
and any discussion of temporal behavior is now meaningless...you have to work in the time
domain to talk about things that happen in time, such as whether or not the modulated carrier
goes to zero on 100% negative modulation peaks!)

73,
Ken W5KFS

[KA1ZGC]

Even though the original question was asked ages ago, and the author has probably moved on to other sites, I'm going to try answering the question that was asked.

The basic question is "why does AM look different when observed from different angles?", the answer is "because you're looking at it from different angles".

When observed mathmatically or through a spectrum analyzer, you see a steady carrier flanked by two opposing sidebands. When observed through an oscilloscope, you see the vector sum of the carrier and sidebands, which appears as a modulated carrier.

The original question asked why you see this modulation on the carrier on a scope and not on a spectrum analyzer. Why don't you see the same unmodulated carrier on a scope that you see on a spectrum analyzer or through mathmatical analysis?

Simple: the scope is not filtered to see only the carrier frequency. It sees the carrier energy and the sideband energy. A heterodyne is a heterodyne no matter where or when it happens, when you combine all these things together, you see a modulated carrier (it works both ways: the carrier heterodyned with the sidebands forms the modulated AM envelope, just as the audio heterodyned with the carrier forms the carrier and sidebands).

The oscilloscope sees the vector sum of the carrier and sideband energy. You're not looking at only the carrier on a scope.

The spectrum analyzer, on the other hand, is very frequency-specific. When it sweeps past the carrier frequency, it (mostly) doesn't detect any of the sideband energy. The carrier itself, without the sidebands to modulate it, appears unmodulated. Mathmatical analysis is even more frequency-specific.

Why doesn't this carrier level shift under modulation during spectral or mathmatical analysis? That too, has a simple answer: because there's no DC component to your voice.

You will see the carrier change amplitude on a spectrum analyzer (or by doing the math) when you observe a modern-day unit generating low-level linear AM and using an ALC designed for SSB. You will see the carrier shrink under modulation, both on a scope and an analyzer. The ALC introduces a DC component that wouldn't be there otherwise.

At any rate, you can't compare scope and analyzer readings. The only thing they have in common is amplitude measurements. Otherwise, they're showing you very different things by very different means. One is a portrait shot, one is a profile shot. One is filtered, one is not.

One is an apple, the other is a dump truck.

You don't see the same thing because you're looking at two different things. Neither piece of equipment is lying, nor is the math. When you consider what is being measured by each unit and how those measurements are taken by each unit, they no longer appear to be mutually-exclusive, and it all makes perfect sense.

[flintstone mop]

I tot that the carrier only disappears in the F.M. mode. Bessel Null (?)
Certain modulating freqs and or amount of deviation result in the disappearance of the carrier.
SSB the carrier is supressed.
PLS clarify me

Fred

[KA1ZGC]

Yes, certain modulating frequencies at certain deviation levels cause carrier cancellation in FM transmissions.

The AM carrier does not "disappear". What you see on your scope is a vector summation of the carrier and both sidebands, it is not filtered down to see only the carrier itself.

[kc6mcw]

Its almost as if you need several viewing layers on the scope so that you can view everything seperately without them adding together. Is this possible?

[KA1ZGC]

That's called a "waterfall" display. It shows you amplitude vs. frequency vs. time.

An oscilloscope can show you amplitude vs. time. A spectrum analyzer can show you amplitude vs. frequency. A waterfall display shows all three dimensions.

You don't "need" that, though. This isn't some shortcoming of a scope. You need to see your carrier and all modulating energy together to determine if you're modulating properly with a scope. That's the only point to even using one. If you filter your scope down to only view energy within a few hertz (just the carrier, or just a portion of just one sideband), the scope pattern will no longer have any real-world meaning.

[kc6mcw]

Thats it! I need a scope with the waterfall display option. I'll check Ebay. HA...But yes, starting to make some sense now.

[Gito]

HI

Maybe I'm wrong but if I change Chas Question like This 

Can I get an RF output when my final amplifier tube has Zero plate voltage on it?

Can I ?,even with a high power driver?


Regards.


Gito

[WB6VHE]

Answer these three questions:  (1) When the modulating signal swings to it's max negative value,
and the modulation percentage is 100%, what is the instantaneous plate voltage of the modulated
rf amplifier?  (2) What is the instantaneous plate current of the modulated amplifier under these same
conditions?
(3) What is the amplitude of the carrier (output of the modulated amplifier) under these conditions?


I'll answer them for you: (1) zero; (2) zero; (3) zero.

Nuff said.  As my son used to say when he was a toddler, "I done!"

[KA1ZGC]

I keep hearing about "instantaneous voltage".

Frequency is a function of time. What happens at a given instant is meaningless without many instants before and after.

You can't do a Fourier transform on a single sample, because there is no frequency information.

At a given frozen instant in time, there is no such thing as radio. Only different voltages on different pieces of matter.

[Steve - WB3HUZ]

None of this mitigates the fact that the carrier is gone at 100% negative peak modulation.

I keep hearing about "instantaneous voltage".

Frequency is a function of time. What happens at a given instant is meaningless without many instants before and after.

You can't do a Fourier transform on a single sample, because there is no frequency information.

At a given frozen instant in time, there is no such thing as radio. Only different voltages on different pieces of matter.

[KA1ZGC]

None of this mitigates the fact that the carrier is gone at 100% negative peak modulation.

It doesn't have to. You can't observe an isolated frozen instant of time and still have any concept of frequency.

[Opcom]

the modulation domain!!
http://www.home.agilent.com/agilent/product.jspx?nid=-35238.536880491.00&cc=US&lc=eng

I forgot about it. but this discussion's clearing alot of things up.

[The Slab Bacon]

Geeezy peezy, Talk about making a mountain out of a mole hill! Do any of youse guys walk around with a shirt that says "kick me" on the back? Dont make it any harder than it is.

There is no carrier at 100% negative, 0, nothing, nada! Just look at your monitor scope, it shows a flat line at 100% negative. It shows high peaks (approximately twice the height of the bare carrier) at 100% positive. You dont need $1,000,000 worth of test gear to see this. A simple scope on the output will show this.

You dont need a degree in nuclear physics to see this. And, er, furthermore............
If you cant determine this from looking at your monitor scope you probably shouldn't be operating this type of gear or posess an amateur lisence! !

A final amplifier tube with no plate voltage makes no RF output of its own. It is just that short, sweet, and simple! ! ! ! ! ! ! ! ! !

                                                                 The Slab Bacon

[KA1ZGC]

If you cant determine this from looking at your monitor scope you probably shouldn't be operating this type of gear or posess an amateur lisence! !

Following that logic, if you can't spell "license", you probably shouldn't have one, either.

So what if your carrier "disappears" for the brief instant you hit 100% negative? That's like saying your carrier "disappears" at the zero crossing: even if it's technically correct, it has no bearing on the practical world.

For that matter, why don't you say that your carrier "disappears" over the 80-some-odd percent of time your grid is driven into the positive, and that the tank circuit doesn't figure into it? That's pretty much the same stance.

[Steve - WB3HUZ]

But it does have practical application. Overmodulation cuts the carrier on and off, thus creating splatter. You are confusing looking at a particular point on a waveform with not looking at the waveform at all. Instantaneous does not mean you completely disregard what comes before or after. Instananeous voltage and current values are used all the time in tube parameters, amplifier design, breakdown testing, etc. That practical applications are many and varied.